n2+9n+18=(n+3)(n+)

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Solution for n2+9n+18=(n+3)(n+) equation: 



n2+9n+18=(n+3)(n+)

We move all terms to the left:

n2+9n+18-((n+3)(n+))=0

We add all the numbers together, and all the variables

n2+9n-((n+3)(+n))+18=0

We add all the numbers together, and all the variables

n^2+9n-((n+3)(+n))+18=0

We multiply parentheses ..

n^2-((+n^2+3n))+9n+18=0



We calculate terms in parentheses: -((+n^2+3n)), so:

(+n^2+3n)

We get rid of parentheses

n^2+3n

Back to the equation:

-(n^2+3n)



We add all the numbers together, and all the variables

n^2+9n-(n^2+3n)+18=0

We get rid of parentheses

n^2-n^2+9n-3n+18=0

We add all the numbers together, and all the variables

6n+18=0

We move all terms containing n to the left, all other terms to the right

6n=-18

n=-18/6

n=-3

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